Bisection Method

Mechanical Engineering

Berhe Nerea Kahsay

Aksum University
College of Natural and Computational
Sciences

1
 Basis of Bisection Method
Theorem An equation f(x)=0, where f(x) is a real continuous
function, has at least one root between xl and xu if
f(xl) f(xu) < 0. f(x)

x
x
xu

Figure 1 At least one root exists between the two points if the
function is real, continuous, and changes sign.

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 Basis of Bisection Method
f(x)

x x
xu

Figure 2 If function f ( x ) does not change sign between two

points, roots of the equation f ( x ) =0 may still exist between the
two points.
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 Basis of Bisection Method
f(x)
f(x)

x xu
x x
x xu

Figure 3 If the function f ( x ) does not change sign between two

points, there may not be any roots for the equation f ( x ) =0
between the two points.

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 Basis of Bisection Method
f(x)

xu
x
x 

Figure 4 If the function f ( x ) changes sign between two points,

more than one root for the equation f ( x ) =0 may exist
between the two points.

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 Step 1
Choose x and xu as two guesses for the root such
that f(x) f(xu) < 0, or in other words, f(x) changes
sign between x and xu. This was demonstrated in
f(x)
Figure 1.

x
x
xu

Figure 1

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 Step 2
Estimate the root, xm of the equation f (x) = 0 as
the mid point between x and xu as
f(x)

x  xu
xm =
2
x xm
x
xu

Figure 5 Estimate of xm
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 Step 3
Now check the following

a) If f ( x l ) f ( x m ) < 0 , then the root lies between x

and xm; then x = x ; xu = xm.

a) If f ( x l ) f ( x m ) > 0 , then the root lies between xm

and xu; then x = xm; xu = xu.

a) If f ( x l ) f ( x m ) =0 ; then the root is xm. Stop the

algorithm if this is true.

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 Step 4
Find the new estimate of the root
x  xu
xm =
2
Find the absolute relative approximate error
old
x new
m
−x m
|∈a|=| new
|100%
xm
where
old
x m =previous estimate of root
new
x m = current estimate of root

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 Step 5
Compare the absolute relative approximate |∈ |
a
error with the pre-specified error tolerance | ∈s |.
Go to Step 2 using
Yes new upper and lower
Is |∈ a|> ∈ s? guesses.

No Stop the algorithm

Note one should also check whether the number

of iterations is more than the maximum number of
iterations allowed. If so, one needs to terminate
the algorithm and notify the user about it.
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 Example 1
You are working for ‘DOWN THE TOILET
COMPANY’ that makes floats for ABC commodes.
The floating ball has a specific gravity of 0.6 and
has a radius of 5.5 cm. You are asked to find the
depth to which the ball is submerged when
floating in water.

Figure 6 Diagram of the floating ball

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 Example 1 Cont.
The equation that gives the depth x to which the
ball is submerged under water is given by
x 3 −0 . 165 x 2 +3 . 993×10−4 =0

a) Use the bisection method of finding roots of

equations to find the depth x to which the ball is
submerged under water. Conduct three iterations
to estimate the root of the above equation.
b) Find the absolute relative approximate error at
the end of each iteration, and the number of
significant digits at least correct at the end of
each iteration.
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 Example 1 Cont.
From the physics of the problem, the ball would
be submerged between x = 0 and x = 2R,
where R = radius of the ball,
that is
0≤x≤2 R
0≤x≤2 ( 0.055 )
0≤x≤0 .11

Figure 6 Diagram of the floating ball

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 Example 1 Cont.
Solution

To aid in the
understanding of how
this method works to find
the root of an equation,
the graph of f(x) is shown
to the right,
where
3 2 −4
f ( x ) =x −0 . 165 x +3 . 993×10
Figure 7 Graph of the function f(x)
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 Example 1 Cont.
Let us assume
x ℓ =0 . 00
x u =0 . 11
Check if the function changes sign between x
and xu .
f ( x l ) =f ( 0 ) =( 0 ) −0 .165 ( 0 ) +3 . 993×10 =3 . 993×10− 4
3 2 −4

f ( x u ) =f ( 0 . 11 ) =( 0 .11 )3 −0 . 165 ( 0 . 11 )2 +3 . 993×10−4 =−2 . 662×10−4

Hence
f ( x l ) f ( x u ) =f ( 0 ) f ( 0 . 11 ) =( 3 . 993×10−4 )(−2 . 662×10−4 ) < 0

So there is at least on root between x and xu, that is between 0

and 0.11
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 Example 1 Cont.

Figure 8 Graph demonstrating sign change between

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initial limits
 Example 1 Cont.
Iteration 1 x ℓ +x u 0+ 0 .11
The estimate of the root is x m= = =0 . 055
2 2
f ( x m ) =f ( 0 .055 )= ( 0 . 055 )3 −0 . 165 ( 0 . 055 )2 +3 . 993×10−4 =6 . 655×10−5
f ( x l ) f ( x m ) =f ( 0 ) f ( 0 .055 )= ( 3 . 993×10− 4 )( 6 .655×10−5 ) >0

Hence the root is bracketed between xm and xu, that is,

between 0.055 and 0.11. So, the lower and upper limits of the
new bracket are
x l =0. 055 , x u =0 .11
At this point, the absolute relative approximate error |∈a|
cannot be calculated as we do not have a previous
approximation.
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 Example 1 Cont.

Figure 9 Estimate of the root for Iteration 1

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 Example 1 Cont.
Iteration 2 x ℓ +x u 0 . 055+ 0 .11
The estimate of the root is x m= = =0 . 0825
2 2
f ( x m ) =f ( 0 .0825 )= ( 0 . 0825 )3 −0 . 165 ( 0 . 0825 ) 2 +3 . 993×10−4 =−1 . 622×10−4
f ( x l ) f ( x m ) =f ( 0 . 055 ) f (0 . 0825 )=( −1. 622×10− 4 ) ( 6 . 655×10−5 ) <0

Hence the root is bracketed between xand xm, that is,

between 0.055 and 0.0825. So, the lower and upper limits of
the new bracket are
x l =0. 055 , x u =0. 0825

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 Example 1 Cont.

Figure 10 Estimate of the root for Iteration 2

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 Example 1 Cont.
The absolute relative approximate error |∈a| at the end of
Iteration 2 is
x new − x old
|a|=| mnew m |×100
xm
0 . 0825−0 . 055
| |×100
0 . 0825
33 . 333%
None of the significant digits are at least correct in the estimate
root of xm = 0.0825 because the absolute relative approximate
error is greater than 5%.

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 Example 1 Cont.
Iteration 3 x ℓ +x u
The estimate of the root is 0 . 055+ 0 . 0825
x m= = =0 . 06875
2 2
f ( x m ) =f ( 0 .06875 )= ( 0 . 06875 )3 −0 . 165 ( 0 . 06875 ) 2 +3 . 993×10−4 =−5 . 563×10−5
f ( x l ) f ( x m ) =f ( 0 . 055 ) f ( 0 . 06875 )=( 6 . 655×10−5 )(−5 . 563×10−5 ) <0

Hence the root is bracketed between xand xm, that is,

between 0.055 and 0.06875. So, the lower and upper limits of
the new bracket are

x l =0. 055 , x u =0 . 06875

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 Example 1 Cont.

Figure 11 Estimate of the root for Iteration 3

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 Example 1 Cont.
The absolute relative approximate error |∈a|at the end of
Iteration 3 is
x new − x old
|a|=| mnew m |×100
xm
0 . 06875−0 . 0825
=| |×100
0 . 06875
=20%
Still none of the significant digits are at least correct in the
estimated root of the equation as the absolute relative
approximate error is greater than 5%.
Seven more iterations were conducted and these iterations are
shown in Table 1.
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 Table 1 Cont.
Table 1 Root of f(x)=0 as function of number of iterations for
bisection method.
Iteration x xu xm a % f(xm)

1 0.00000 0.11 0.055 ---------- 6.655×10−5

2 0.055 0.11 0.0825 33.33 −1.622×10−4
3 0.055 0.0825 0.06875 20.00 −5.563×10−5
4 0.055 0.06875 0.06188 11.11 4.484×10−6
5 0.06188 0.06875 0.06531 5.263 −2.593×10−5
6 0.06188 0.06531 0.06359 2.702 −1.0804×10−5
7 0.06188 0.06359 0.06273 1.370 −3.176×10−6
8 0.06188 0.06273 0.0623 0.6897 6.497×10−7
9 0.0623 0.06273 0.06252 0.3436 −1.265×10−6
10 0.0623 0.06252 0.06241 0.1721 −3.0768×10−7

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 Table 1 Cont.
Hence the number of significant digits at least correct is given
by the largest value or m for which
|a|≤0 . 5×102−m
0 . 1721≤0 . 5×102−m
0 . 3442≤10 2−m
log ( 0 . 3442 )≤2−m
m≤2−log ( 0 .3442 )=2 . 463
So
m= 2
The number of significant digits at least correct in the
estimated root of 0.06241 at the end of the 10th iteration is 2.

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 Advantages
 Always convergent
 The root bracket gets halved with
each iteration - guaranteed.

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 Drawbacks
 Slow convergence
 If one of the initial guesses is close
to the root, the convergence is
slower

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 Drawbacks (continued)
 If a function f(x) is such that it just
touches the x-axis it will be unable
to find the lower and upper guesses.

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 Drawbacks (continued)
 Function changes sign but root does
not exist

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